Solve 800x^2+800x-1500=0 | Microsoft Math Solver (2024)

800x^{2}+800x-1500=0

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

x=\frac{-800±\sqrt{800^{2}-4\times 800\left(-1500\right)}}{2\times 800}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 800 for a, 800 for b, and -1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-800±\sqrt{640000-4\times 800\left(-1500\right)}}{2\times 800}

Square 800.

x=\frac{-800±\sqrt{640000-3200\left(-1500\right)}}{2\times 800}

Multiply -4 times 800.

x=\frac{-800±\sqrt{640000+4800000}}{2\times 800}

Multiply -3200 times -1500.

x=\frac{-800±\sqrt{5440000}}{2\times 800}

Add 640000 to 4800000.

x=\frac{-800±400\sqrt{34}}{2\times 800}

Take the square root of 5440000.

x=\frac{-800±400\sqrt{34}}{1600}

Multiply 2 times 800.

x=\frac{400\sqrt{34}-800}{1600}

Now solve the equation x=\frac{-800±400\sqrt{34}}{1600} when ± is plus. Add -800 to 400\sqrt{34}.

x=\frac{\sqrt{34}}{4}-\frac{1}{2}

Divide -800+400\sqrt{34} by 1600.

x=\frac{-400\sqrt{34}-800}{1600}

Now solve the equation x=\frac{-800±400\sqrt{34}}{1600} when ± is minus. Subtract 400\sqrt{34} from -800.

x=-\frac{\sqrt{34}}{4}-\frac{1}{2}

Divide -800-400\sqrt{34} by 1600.

x=\frac{\sqrt{34}}{4}-\frac{1}{2} x=-\frac{\sqrt{34}}{4}-\frac{1}{2}

The equation is now solved.

800x^{2}+800x-1500=0

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

800x^{2}+800x-1500-\left(-1500\right)=-\left(-1500\right)

Add 1500 to both sides of the equation.

800x^{2}+800x=-\left(-1500\right)

Subtracting -1500 from itself leaves 0.

800x^{2}+800x=1500

Subtract -1500 from 0.

\frac{800x^{2}+800x}{800}=\frac{1500}{800}

Divide both sides by 800.

x^{2}+\frac{800}{800}x=\frac{1500}{800}

Dividing by 800 undoes the multiplication by 800.

x^{2}+x=\frac{1500}{800}

Divide 800 by 800.

x^{2}+x=\frac{15}{8}

Reduce the fraction \frac{1500}{800} to lowest terms by extracting and canceling out 100.

x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{15}{8}+\left(\frac{1}{2}\right)^{2}

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}+x+\frac{1}{4}=\frac{15}{8}+\frac{1}{4}

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

x^{2}+x+\frac{1}{4}=\frac{17}{8}

Add \frac{15}{8} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

\left(x+\frac{1}{2}\right)^{2}=\frac{17}{8}

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{8}}

Take the square root of both sides of the equation.

x+\frac{1}{2}=\frac{\sqrt{34}}{4} x+\frac{1}{2}=-\frac{\sqrt{34}}{4}

Simplify.

x=\frac{\sqrt{34}}{4}-\frac{1}{2} x=-\frac{\sqrt{34}}{4}-\frac{1}{2}

Subtract \frac{1}{2} from both sides of the equation.

x ^ 2 +1x -\frac{15}{8} = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 800

r + s = -1 rs = -\frac{15}{8}

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = -\frac{1}{2} - u s = -\frac{1}{2} + u

Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -\frac{15}{8}

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{15}{8}

\frac{1}{4} - u^2 = -\frac{15}{8}

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -\frac{15}{8}-\frac{1}{4} = -\frac{17}{8}

Simplify the expression by subtracting \frac{1}{4} on both sides

u^2 = \frac{17}{8} u = \pm\sqrt{\frac{17}{8}} = \pm \frac{\sqrt{17}}{\sqrt{8}}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{1}{2} - \frac{\sqrt{17}}{\sqrt{8}} = -1.958 s = -\frac{1}{2} + \frac{\sqrt{17}}{\sqrt{8}} = 0.958

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.